Gases
CHAPTER 10
• Powerpoint from Melissa Brophy’s website
(http://teacherweb.com/TX/McNeilHS/brophy/
photo7.aspx) and modified for our class
Resources and Activities
• Textbook - chapter 10 & ppt file
• Lab activities
(1) Exploring Gas Behavior
(with LabQuest Units )
(2) Molar volume of a gas (wet lab)
• POGIL activities:
Ideal gas law
Partial Pressure
• Chemtour videos from W.W. Norton
(chapter 6 : ideal gas law; Dalton’s law;
molecular speed)
• http://www.wwnorton.com/college/chemistry
/chemistry3/ch/06/chemtours.aspx
Gases
(Chapter 10)
Animation on gases to view:
http://glencoe.com/sites/common_assets/advanced_placement/chemistry_
chang9e/animations/chang_7e_esp/gam2s2_6.swf
Independent work - view animations & interactive activities (4 in total) and
write summary notes on each. These summaries are to be included in your
portfolio
http://glencoe.mcgrawhill.com/sites/0035715985/student_view0/chapter5/animations_center.html#
Activities and Problem set for chapter 10
(due date_______)
TextBook ch. 10– all sections required
for regents (in part), SAT II and AP
exams
•
Complete the chapter 10 packet
part (a) Characteristics of Gases
( sample, practice and integrative
exercises - show work) and part (b)
Lab activities: (2)
POGILS (2)
AP Chemistry Gas Law Practice
•
Do all chapter 10 GIST and
Visualizing concepts problems write out questions and answers &
show work. (NB Photocopy of
questions is also acceptable)
Measurements on Gases
•
Properties of gases
–
–
–
–
Gases uniformly fill any container
Gases are easily compressed
Gases mix completely with any other gas
Gases exert pressure on their surroundings
•
Pressure = Force
Area
Measurements on Gases
•
Measuring pressure
–
The barometer –
measures atmospheric
pressure
•
Inventor – Evangelista
Torricelli (1643)
Measurements on Gases
– The manometer – measures confined gas
pressure
Measurements on Gases
–
Units
•
mm Hg (torr)
–
•
Kilopascal (kPa)
–
•
101.325 kPa = standard pressure
Atmospheres
–
•
760 torr = standard pressure
1 atmosphere (atm) = standard pressure
STP = 1 atm = 760 torr = 760 mmHg = 101.325 kPa
Measurements on Gases
• Example: Convert 0.985 atm to torr and to kPa.
0.985 atm
760 torr
1 atm
0.985 atm
101.325 kPa
1 atm
= 749 torr
= 98.3 kPa
The Gas Laws of Boyle, Charles and Avogadro
• Boyle’s Law (Robert Boyle, 16271691)
– The product of pressure times volume is
a constant, provided the temperature
remains the same
• PV = k
The Gas Laws of Boyle, Charles and Avogadro
•
•
•
P is inversely related to V
The graph of P versus V is hyperbolic
Volume increases linearly as the pressure
decreases
– As you squeeze a zip lock bag filled with air
(reducing the volume), the pressure increases
making it difficult to keep squeezing
The Gas Laws of Boyle, Charles and Avogadro
–
At constant temperature, Boyle’s law can be
used to find a new volume or a new pressure
•
P1V1 = k = P2V2
or P1V1 = P2V2
–
–
Boyle’s law works best at low pressures
Gases that obey Boyle’s law are called Ideal
gases
The Gas Laws of Boyle, Charles and Avogadro
• Example: A gas which has a pressure of 1.3 atm
occupies a volume of 27 L. What volume will the
gas occupy if the pressure is increased to 3.9 atm
at constant temperature?
•
•
•
•
P1 = 1.3 atm
V1 = 27 L
P2 = 3.9 atm
V2 = ?
(1.3atm)(27L) = (3.9atm)V2
V2 = 9.0 L
The Gas Laws of Boyle, Charles and Avogadro
•
Charles’ Law (Jacques Charles, 1746 –
1823)
– The volume of a gas increase linearly with
temperature provided the pressure remains
constant
V = bT
V=b
T
or V1 = V2
T1
T2
V1 = b = V2
T1
T2
The Gas Laws of Boyle, Charles and Avogadro
•
Temperature must be measured in Kelvin
( K = °C + 273)
– 0 K is “absolute zero”
The Gas Laws of Boyle, Charles and Avogadro
• Example: A gas at 30°C and 1.00 atm has a
volume of 0.842L. What volume will the gas
occupy at 60°C and 1.00 atm?
• V1 = 0.842L
• T1 = 30°C (+273 = 303K)
• V2 = ?
• T2 = 60°C (+273 = 333K)
0.842L
303K
V2 = 0.925L
=
V2
333K
The Gas Laws of Boyle, Charles and Avogadro
•
Avogadro’s Law (Amedeo Avogadro, 1811)
–
For a gas at constant temperature and
pressure, the volume is directly proportional to
the number of moles, n
•
V = an
V1 = a = V2
n1
n2
V/n = a
or
V1 = V2
n1
n2
The Gas Laws of Boyle, Charles and Avogadro
• Example: A 5.20L sample at 18°C and 2.00
atm contains 0.436 moles of a gas. If we
add an additional 1.27 moles of the gas at
the same temperature and pressure, what
will the total volume occupied by the gas be?
• V1 = 5.20L
• n1 = 0.436 mol
• V2 = ?
• n2 = 1.27 mol + 0.436 mol = 1.706 mol
The Gas Laws of Boyle, Charles and Avogadro
5.20 L
=
V2
0.436 mol 1.706 mol
x = 20.3L
The Ideal Gas Law
•
Derivation from existing laws
V=k
V=bT
V=an
P
V = kba(Tn)
P
The Ideal Gas Law
Constants k, b, a are combined into the universal
gas constant (R),
V = nRT
P
or
PV = nRT
The Ideal Gas Law
R = PV using standard numbers will give you R
nT
STP
• P = 1 atm
• V = 22.4L
• n = 1 mol
• T = 273K
• Therefore if we solve for R, R = 0.0821 L• atm/mol •K
The Ideal Gas Law
• Example: A sample containing 0.614 moles
of a gas at 12°C occupies a volume of 12.9L.
What pressure does the gas exert?
P=?
V = 12.9L
n = 0.614 mol
R = 0.0821 L•atm/mol• K
T = 12 + 273 = 285K
The Ideal Gas Law
(P)(12.9L) = (0.614mol)(0.0821 L•atm/mol• K)(285K)
Hint: Rearrange and re-write (watch out for units in numerator and denominator)
P =1.11 atm
The Ideal Gas Law
Solving for new volumes, temperature, or pressure
(n remaining constant)
•
Combined Law
P1V1 = nR = P2V2
T1
T2
or P1V1 = P2V2
T1
T2
The Ideal Gas Law
• Example: A sample of methane gas at
0.848 atm and 4.0°C occupies a volume of
7.0L. What volume will the gas occupy if the
pressure is increased to 1.52 atm and the
temperature increased to 11.0°C?
• P1 = 0.848 atm
P2 = 1.52 atm
• V1 = 7.0L
V2 = ?
• T1 = 4+273 = 277K
T2 = 11+273 = 284K
The Ideal Gas Law
(0.848atm)(7.0L)
277K
V2 = 4.0L
= (1.52atm)(V2)
284K
Gas Stoichiometry
•
Molar volume
–
•
One mole of an ideal gas occupies 22.4L of volume at
STP
Things to remember
–
Density = mass
–
volume
n = grams of substance = m
molar mass
M
•
PV = mRT
M
Gas Stoichiometry
• Example: A sample containing 15.0g of dry
ice (CO2(s)) is put into a balloon and allowed
to sublime according to the following
equation: CO2(s)  CO2(g)
How big will the balloon be (i.e. what is the
volume of the balloon) at 22°C and 1.04 atm,
after all of the dry ice has sublimed?
Gas Stoichiometry
• Moles of CO2(s):
15g 1 mol = 0.341 mol CO2(s)
44 g
0.341 mol CO2(s) 1 CO2(g) = 0.341 mol CO2(g)
1 CO2(s)
=n
Gas Stoichiometry
P = 1.04 atm
V=?
n = 0.341 mol
R = 0.0821 L•atm/mol• K
T = 22C + 273 = 295K
V = 7.94L
Gas Stoichiometry
• Limiting Reactant Example: 0.500L of
H2(g) are reacted with 0.600L of O2(g) at
STP according to the equation 2H2(g) +
O2(g)  2H2O(g)
• What volume will the H2O occupy at 1.00 atm
and 350°C?
Gas Stoichiometry
• H2: 0.5L 1 mol 2 H2O = 0.0223 mol H2O
22.4L 2 H2
• O2: 0.6L 1 mol 2 H2O = 0.0536 mol H2O
22.4L 1 O2
Gas Stoichiometry
• H2 :
(1 atm)(V) = (0.0223mol)(0.0821 L•atm/mol• K)(623K)
V = 1.14L
Limiting reactant and how
much volume is produced!
• O 2:
(1atm)(V) = (0.0536mol)(0.0821 L•atm/mol• K)(623K)
V = 2.74L
Gas Stoichiometry
• Density/Molar Mass example: A gas at 34°C and
1.75 atm has a density of 3.40g/L. Calculate the
molar mass of the gas.
D = mass
3.40 = 3.40g
V
1L
P = 1.75 atm
V = 1L
m = 3.40 g
R = 0.0821 L•atm/mol• K
T = 34 + 273 = 307K
M=x
Gas Stoichiometry
(1.75atm)(1L) = (3.4g)(0.0821 L•atm/mol• K)(307K)
M
M = molar mass
M = 48.97 g/mol
Dalton’s Law of Partial Pressures (John Dalton, 1803)
•
Statement of law
– “for a mixture of gases in a container, the total
pressure exerted is the sum of the pressures
each gas would exert if it were alone”
– It is the total number of moles of particles that
is important, not the identity or composition of
the gas particles.
Dalton’s Law of Partial Pressures
•
Derivation
–
–
–
–
–
Ptotal = P1 + P2 + P3 …
P1 = n1RT
P2 = n2RT
P3 = n3RT ….
V
V
V
Ptotal = n1RT + n2RT + n3RT
V
V
V
Ptotal = (n1+n2+n3+…) (RT)
V
Ptotal = ntotal (RT)
V
Dalton’s Law of Partial Pressures
• Example: Oxygen gas is collected over
water at 28C. The total pressure of the
sample is 5.5 atm. At 28C, the vapor
pressure of water is 1.2 atm. What pressure
is the oxygen gas exerting?
• Ptotal = PO2 + PH2O
• 5.5 = x + 1.2
• X = 4.3 atm
Dalton’s Law of Partial Pressures
•
Mole Fraction
– The ratio of the number of moles of a given
component in a mixture to the total number of
moles in the mixture
– For an ideal gas, the mole fraction (x):
x1 = n1
=
ntotal
P1
Ptotal
Dalton’s Law of Partial Pressures
• Example: The vapor pressure of water in air
at 28°C is 28.3 torr. Calculate the mole
fraction of water in a sample of air at 28°C
and 1.03 atm.
• XH2O = PH2O = 28.3 torr
Pair
783 torr
= 0.036
Dalton’s Law of Partial Pressures
• Example: A mixture of gases contains 1.5
moles of oxygen, 7.5 moles of nitrogen and
0.5 moles of carbon dioxide. If the total
pressure exerted is 800 mmHg, what are the
partial pressures of each gas in the mixture?
Dalton’s Law of Partial Pressures
• Calculate the total number of moles
– 1.5 + 7.5 + 0.5 = 9.5 moles
• Use mole fractions for each individual gas
• PO2: 800mmHg x 1.5 mol = 126 mmHg
9.5 mol
• PN2: 800mmHg x 7.5 mol = 632 mmHg
9.5 mol
• PCO2: 800mmHg x 0.5 mol = 42 mmHg
9.5 mol
Kinetic Molecular Theory of Gases
•
Postulates of the KMT Related to Ideal Gases
–
–
–
–
The particles are so small compared with the distances
between them that the volume of the individual particles
can be assumed to be zero
The particles are in constant motion. Collisions of the
particles with the walls of the container cause pressure.
Assume that the particles exert no forces on each other
The average kinetic energy of a collection of gas
particles is assumed to be directly proportional to the
Kelvin temperature of the gas
Kinetic Molecular Theory of Gases
•
Explaining Observed Behavior with KMT
– P and V (T = constant)
•
•
As V is decreased, P increases:
V decrease causes a decrease in the
surface area. Since P is force/area, the
decrease in V causes the area to decrease,
thus increasing the P
Kinetic Molecular Theory of Gases
– P and T (V = constant)
•
•
As T increase, P increases
The increase in T causes an increase in
average kinetic energy. Molecules moving
faster collide with the walls of the container
more frequently, and with greater force.
Kinetic Molecular Theory of Gases
– V and T (P = constant)
•
•
As T increases, V also increases
Increased T creates more frequent, more
forceful collisions. V must increase
proportionally to increase the surface area,
and maintain P
Kinetic Molecular Theory of Gases
– V and n (T and P constant)
•
•
As n increases, V must increase
Increasing the number of particles increases
the number of collisions. This can be
balanced by an increase in V to maintain
constant P
Kinetic Molecular Theory of Gases
– Dalton’s Law of partial pressures
•
•
P is independent of the type of gas molecule
KMT states that particles are independent,
and V is assumed to be zero. The identity of
the molecule is therefore unimportant
Kinetic Molecular Theory of Gases
•
Root Mean Square Velocity
–
Velocity of a gas is dependent on mass and
temperature
Velocity of gases is determined as an average
–
•
•
M = mass of one mole of gas particles in kg
R = 8.3145 J/K•mol
–
–
joule = kg•m2/s2
urms = 3RT
M
Kinetic Molecular Theory of Gases
• Example: Calculate the root mean square
velocity for the atoms in a sample of oxygen
gas at 0°C.
• MO2 = 32g/mol
• 32g 1 kg = 0.032 kg
1000g
• T = 0 + 273 = 273K
• R = 8.3145 J/K•mol
Kinetic Molecular Theory of Gases
• Urms =
3(8.3145 kg•m2 •s-2•K-1 •mol-1)(273 K) = 461m•s-1
0.032 kg•mol-1
Answer 461m•s-1 or 461 m/s
Recall 1 Joule = 1 J = 1 kg• m2/s2 = 1 kg• m2• s-2
Kinetic Molecular Theory of Gases
•
Mean Free Path – distance between
collisions
–
Average distance a molecule travels between
collisions
•
1 x 10 -7 m for O2 at STP
Effusion and Diffusion
•
Effusion
–
–
Movement of a gas through a small opening
into an evacuated container (vacuum)
Graham’s law of effusion
Rate of effusion for gas 1 = √M2
Rate of effusion for gas 2 √M1
Effusion and Diffusion
• Example: How many times faster than He
would NO2 gas effuse?
• MNO2 = 46.01 g/mol
larger mass
• MHe = 4.003 g/mol
effuses slower
• Rate He =
Rate NO2
46.01
4.003
=
He effuses 3.39
times faster
Real Gases and van der Waals Equation
(Johannes van der Waals, 1873)
•
Volume
–
–
Real gas molecules do have volume
Volume available is not 100% of the container
volume
•
•
n= number of moles
b = is an empirical constant, derived from
experimental results
Real Gases and van der Waals Equation
• Ideal: P = nRT
V
• Real: P = nRT
V-nb
Real Gases and van der Waals Equation
•
Pressure
–
Molecules of real gases do experience
attractive forces
•
•
a = proportionality constant determined by
observation of the gas
Pobs = Pideal – a(n/V)2
Real Gases and van der Waals Equation
•
•
Combining to derive van der Waals
equation
Pobs = nRT - n2a
V-nb V2
And then rearranging…
• (Pobs + n2a/V2)(V-nb) = nRT
Chemistry in the Atmosphere
•
Composition of the Troposphere
– The atmosphere is composed of 78% N2, 21%
O2, 0.9%Ar, and 0.03% CO2 along with trace
gases.
– The composition of the atmosphere varies as a
function of distance from the earth’s surface.
Heavier molecules tend to be near the surface
due to gravity.
Chemistry in the Atmosphere
– Upper atmospheric chemistry is largely affected
by ultraviolet, x ray, and cosmic radiation
emanating from space. The ozone layer is
especially reactive to ultraviolet radiation
– Manufacturing and other processes of our
modern society affect the chemistry of our
atmosphere. Air pollution is a direct result of
such processes.